Question

A spring of spring constant k is attached to a support at the bottom of a ramp that makes an angle θ with the horizontal. A block of inertia m is pressed against the free end of the spring until the spring is compressed a distance d from its relaxed length. Call this position A. The block is then released and moves up the ramp until coming to rest at position B. The surface is rough from position A for a distance 2d up the ramp, and over this distance the coefficient of kinetic friction for the two surfaces is μ . Friction is negligible elsewhere.
What is the distance from A to B? Suppose the values of μ, k, d, θ, and m are such that the spring fully extends, but the block never goes higher than 2d .
Express your answer in terms of some or all of the variables k, μ, m, d, and θ.


More than anything I'm just confused by the question. If someone could at least draw the picture I might be able to figure it out. Thanks.

Answer 1

Answer:

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

Explanation:

From the law of conservation of energy

Energy lost  by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr

$0.5kd^{2}=0.5mv^{2}+mgLsin\theta+\mu_{k}mgcos\theta x$

$x(mgsin\theta+\mu_{k}mgcos\theta)=0.5kd^{2}$

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

The required distance from A to B is $x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$